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JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Matrices

If

X = x y z

is a solution of the system of equations

A X = B

, where

adj A = 4 - 5 1 20 - 2 253

and

B = 402

, then

∣ x + y + z ∣

is equal to :

(A)3
(B) $\frac{3}{2}$
(C)1
(D)2✓ Correct

Answer: (D)

Solution

Since

A (adj A) = ∣ A ∣ I

and

det (adj A) = ∣ A ∣^{2}

for a

3 \times 3

matrix, compute

det (adj A) = 100

, so

∣ A ∣ = \pm 10

. Then

X = \frac{1}{∣ A ∣} (adj A) B = \frac{1}{\pm 10} 4 - 5 1 20 - 2 253 402 = \frac{1}{\pm 10} 20 - 10 10 = \pm 2 - 1 1 .

Therefore

x + y + z = \pm 2

, and

∣ x + y + z ∣ = 2

.

JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Single correct option

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